array values in consecutive order

Question

I build up an array with a counter when a function is true.

So if it were true 3 times in a row, the array looks like [1,2,3]. If the function is not true there is a gap in the counter and could look like this [1,2,3,5].

In another function I need to determine if the array length is > 2 and the values in the array are in consecutive order. So [1,2,3] it should return true. If [1,2,3,5] it should return false.

I haven't found anything that's worked. Any help with a possible solution would be much appreciated.

I have seen this (and have tried it) but it doesn't work.

Array.prototype.is_consecutive = (function () {
    var offset = 0; // remember the last offset
    return function () {
        var start = offset, len = this.length;
        for (var i = start + 1; i < len; i++) {
            if (this[i] !== this[i - 1] + 1) {
                break;
            }
        }
        offset = i;
        return this[start];
    };
})();

Were have you found that?and copy + paste does not count as a try — Jonas Wilms
– Jonas Wilms, Commented Mar 16, 2018 at 20:29
Jonas not a copy and paste. Got it off this site, can't find it now. Thought it was working when I built a temporary jsfiddle for it; but can't get my head around a fix for it. — Rob
– Rob, Commented Mar 16, 2018 at 20:32

Sébastien · Accepted Answer · 2018-03-16 20:36:06Z

2

If you can absolutely rely on the algorithm that populates your array, following the stated rules

I build up an array with a counter when a function is true. So if it were true 3 times in a row, the array looks like [1,2,3]. If the function is not true there is a gap in the counter and could look like this [1,2,3,5].

Then you all you need to do is check that the last element of the array is the same value as the length of the array:

var good = [1, 2, 3];
var bad = [1, 2, 3, 5];

var isValid = function(arr) {
  return (arr.length > 2 && arr[arr.length - 1] === arr.length); // Thanks to Jonas W.
}
console.log(isValid(good));
console.log(isValid(bad));

edited Mar 16, 2018 at 20:36

answered Mar 16, 2018 at 20:32

Sébastien

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2 Comments

Abr001am Over a year ago

what if a value is duplicated ? like this bad=[1,2,2,4] ?

Sébastien Over a year ago

That's not how the algorithm is supposed to work. (read the question) The answer is only appropriate for this specific question.

Nina Scholz · Accepted Answer · 2018-03-16 20:41:39Z

1

You could use a closure over the first index value and increment that value while checking.

function isConsecutive(array) {
    return array.length >= 2 && array.every((v => a => a === v++)(array[0]));
}

console.log([[1, 2, 3, 4], [1, 2, 3, 5]].map(isConsecutive));

edited Mar 16, 2018 at 20:41

answered Mar 16, 2018 at 20:33

Nina Scholz

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Comments

Luis felipe De jesus Munoz · Accepted Answer · 2018-03-16 20:37:40Z

0

var array=[5,6,7,8];

function continuous(arr){
  var max = Math.max(...arr);
  var min = Math.min(...arr);

  return (((max-min)+1) * (min + max) / 2) == arr.reduce((a, b) => a + b, 0)
}

console.log(continuous(array))

answered Mar 16, 2018 at 20:37

Luis felipe De jesus Munoz

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Jonas Wilms · Accepted Answer · 2018-03-16 20:38:50Z

0

 Array.prototype.isConsecutive = function(){
   for(var i = 1; i < this.length; i++)
     if(this[i] !== this[i - 1] + 1)
        return false;
   return true;
 };

Or the obfuscated version:

  const isConsecutive = arr => !!arr.reduce((prev, curr) => curr === prev + 1 ? curr : 0);

edited Mar 16, 2018 at 20:38

answered Mar 16, 2018 at 20:33

Jonas Wilms

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Abr001am · Accepted Answer · 2018-03-17 01:08:13Z

0

You can use this too, it is recursive:

Array.prototype.isConsecutive = function(){
if (arguments.length==0) return this.isConsecutive(0);
else if (arguments[0]>=this.length) return true;
else return (this[arguments[0]]==arguments[0]+1&&this.isConsecutive(arguments[0]+1
));}
cons=[1,2,3];
console.log(cons.isConsecutive());
non_cons=[1,2,2,4];
console.log(non_cons.isConsecutive());

answered Mar 17, 2018 at 1:08

Abr001am

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